CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set 11

Surface Areas and Volumes

Question. A rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is :
(a) 3.5
(b) 7
(c) 2
(d) 5
Answer: (a) 3.5
Explanation :
Given, Area of the sheet of paper \( = (40 \times 22) \text{ cm}^2 = 880 \text{ cm}^2 \)
Height of the cylinder \( = 40 \text{ cm} \)
Let \( r \) be the radius of the cylinder.
Now, curved surface area of the cylinder \( = 2\pi rh \)
Thus, \( 2\left(\frac{22}{7}\right)r(40) = 40 \times 22 \)
\( \Rightarrow 2\left(\frac{1}{7}\right)r = 1 \times 1 \)
\( \Rightarrow r = \frac{7}{2} \text{ cm} = 3.5 \text{ cm} \)

Question. The number of solid spheres each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm are :
(a) 3
(b) 5
(c) 4
(d) 6
Answer: (b) 5
Explanation :
Given, Height of cylinder \( = 45 \text{ cm} \)
Diameter of cylinder \( = 4 \text{ cm} \)
Diameter of each sphere \( = 6 \text{ cm} \)
Thus, radius of cylinder \( = 2 \text{ cm} \)
and radius of each sphere \( = 3 \text{ cm} \)
\( \therefore \text{Number of spheres} = \frac{\text{Volume of cylinder}}{\text{Volume of each sphere}} \)
\( = \frac{\pi(2)^2 45}{\frac{4}{3}\pi(3)^3} \)
\( = \frac{4 \times 45}{4 \times 9} = 5 \)

Question. The surface area of two spheres are in the ratio 16 : 9. The ratios of their volumes is :
(a) 64 : 27
(b) 16 : 9
(c) 4 : 3
(d) \( 16^3 : 9^3 \)
Answer: (a) 64 : 27
Explanation :
Let the radius of the larger sphere be R and the smaller be r.
Given, \( 4\pi R^2 : 4\pi r^2 = 16 : 9 \)
\( \Rightarrow R^2 : r^2 = 4^2 : 3^2 \)
\( \Rightarrow R : r = 4 : 3 \)
Thus, \( V_1 : V_2 = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = 4^3 : 3^3 \)
or \( V_1 : V_2 = 64 : 27 \)

Question. A solid right-circular cone is cut into two parts at the middle of its height by a plane parallel to its base. The ratio of the volume of the smaller cone to the whole cone is :
(a) 1 : 2
(b) 1 : 4
(c) 1 : 6
(d) 1 : 8
Answer: (d) 1 : 8
Explanation :
Let the height of the bigger cone be \( 2x \text{ cm} \).
Thus, height of the smaller cone \( = x \text{ cm} \)
Also let the radius of the larger cone be \( 2r \text{ cm} \).
Thus, radius of the smaller cone \( = r \text{ cm} \) (by mid-point theorem)
Now, volume of the large cone \( V = \frac{1}{3} \times \pi(2r)^2 (2x) = \frac{8\pi}{3} r^2 x \)
and volume of the smaller cone \( = \frac{1}{3}\pi(r)^2 (x) = \frac{\pi}{3} r^2 x \)
Thus, \( \frac{\text{Volume of the smaller cone}}{\text{Volume of the larger cone}} = \frac{\frac{\pi}{3} r^2 x}{\frac{8\pi}{3} r^2 x} = \frac{1}{8} \)

Question. If the radius of the base of a right-circular cylinder is halved keeping the height same then the ratio of the volume of the cylinder thus obtained to the volume of the original cylinder is :
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer: (c) 1 : 4
Explanation :
Sol. Let the original radius of the base be \( 2r \text{ cm} \) and the height be \( h \).
Thus, the new radius of the base \( = r \text{ cm} \)
Now, Original volume \( = \pi(2r)^2 h \)
New volume \( = \pi(r)^2 h \)
Thus, \( \frac{\text{Original volume}}{\text{New volume}} = \frac{\pi \times 4r^2 h}{\pi r^2 h} = \frac{4}{1} \)
\( \Rightarrow \frac{\text{New volume}}{\text{Original volume}} = \frac{1}{4} \)

Question. The radius of the largest circular cone that can be cut out from a cube of edge 4.2 cm is :
(a) 4.2 cm
(b) 2.1 cm
(c) 8.4 cm
(d) 1.05 cm
Answer: (b) 2.1 cm
Explanation :
Given, the edge of cube \( = 4.2 \text{ cm} \)
Now, the largest circular cone will have diameter equal to the measure of the edge.
Thus, diameter \( = 4.2 \text{ cm} \)
Hence, radius \( = 2.1 \text{ cm} \)

Question. A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm partly filled with water. If the sphere is completely submerged then the water level rises (in cm) by :
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (a) 3
Explanation :
Given, Diameter of sphere \( = 18 \text{ cm} \)
Diameter of cylinder \( = 36 \text{ cm} \)
Let the increase in height of water in the cylinder be \( h \text{ cm} \).
Radius of sphere \( = 9 \text{ cm} \)
Radius of cylinder \( = 18 \text{ cm} \)
Thus, volume of sphere \( = \frac{4}{3}\pi(9)^3 \text{ cm}^3 \)
\( = \frac{4}{3}\pi \times (81) (9) \text{ cm}^3 = 81 \times 12\pi \text{ cm}^3 \)
Thus, increase in volume of water level in the cylinder \( = \pi (18)^2 h \text{ cm}^3 \)
Hence, \( \pi(18)^2 h = (81)(12)\pi \)
\( \Rightarrow 4h = 12 \)
\( \Rightarrow h = 3 \text{ cm} \)

Question. A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface area of the two are equal then the ratio of the radius and the slant height of the conical part is :
(a) 2 : 1
(b) 1 : 2
(c) 1 : 4
(d) 4 : 1
Answer: (b) 1 : 2
Explanation :
Given, Radius of hemisphere \( = \text{Radius of cone} \)
Also, curved surface area of hemisphere \( = \text{Curved surface area of cone} \)
\( \Rightarrow 2\pi r^2 = \pi rl \) [where \( l = \text{slant height of the cone} \)]
\( \Rightarrow 2r = l \)
\( \Rightarrow r : l = 1 : 2 \)

Question. A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. If the total height of the toy is 31 cm, then the height of the cone is :
(a) 31 cm
(b) 38 cm
(c) 7 cm
(d) 24 cm
Answer: (d) 24 cm
Explanation :
Radius of sphere \( = \text{Radius of cone} \)
\ Height of cone \( = \text{Total height of cone} - \text{Radius of sphere} \)
\( = (31 - 7)\text{cm} = 24 \text{ cm} \)

Question. A solid sphere of radius r is melted and casted into the shape of a solid cone of height r, then the radius of the cone’s base is :
(a) 2r
(b) 3r
(c) 4r
(d) 6r
Answer: (a) 2r
Explanation :
Radius of sphere \( = r \)
\ Volume of sphere \( = \frac{4}{3}\pi r^3 \)
Radius of the base of cone be R and height, \( h = r \).
\ Volume of cone \( = \frac{1}{3}\pi R^2 r \)
But volume of cone \( = \text{Volume of sphere} \)
\( \Rightarrow \frac{1}{3}\pi R^2 r = \frac{4}{3}\pi r^3 \)
\( \Rightarrow R^2 = 4r^2 \)
\( \Rightarrow R = 2r \)

Question. Three cubes each of side 12 cm are joined end to end. The total surface area of the resulting solid is :
(a) 2016 cm²
(b) 3150 cm²
(c) 1575 cm²
(d) 1008 cm²
Answer: (a) 2016 cm²
Explanation :
Then, \( \text{Length} = 3l \Rightarrow 3 \times 12 = 36 \)
\( \text{Breadth, } b = 12 \)
\( \text{Height, } h = 12 \)
Total surface area \( = 2(lb + bh + hl) \)
\( = 2(36 \times 12 + 12 \times 12 + 12 \times 36) \)
\( = 2(432 + 144 + 432) \)
\( = 2016 \text{ cm}^2 \)

Question. Volumes of two spheres are in the ratio 125 : 216. The ratio of their surface areas is :
(a) 5 : 6
(b) 25 : 36
(c) 1 : 2
(d) 5 : 2
Answer: (b) 25 : 36
Explanation :
Volume of sphere \( = \frac{4}{3}\pi r^3 \)
Let radius of sphere be \( r_1 \) and \( r_2 \) so
\( \text{Volume } (V_1) = \frac{4}{3}\pi r_1^3 \) and \( \text{Volume } (V_2) = \frac{4}{3}\pi r_2^3 \)
\( \Rightarrow \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{125}{216} \)
\( \Rightarrow \frac{r_1}{r_2} = \frac{5}{6} \dots \text{(i)} \)
Ratio of surface area \( = \frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 \)
\( = \left(\frac{5}{6}\right)^2 \) [From (i)]
\ Ratio of surface area \( = \frac{25}{36} \)

Question. Volume of a metallic sphere of radius 3 cm is :
(a) 36π cm³
(b) 108π cm³
(c) 54π cm³
(d) 6π cm³
Answer: (a) 36π cm³
Explanation :
Volume of metallic sphere \( = \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times \pi \times 3 \times 3 \times 3 \)
\( = 36\pi \text{ cm}^3 \)

Question. The ratio between volume of two spheres is 8 : 27. What will be ratio between their surface areas?
(a) 4 : 9
(b) 4 : 5
(c) 2 : 1
(d) 5 : 3
Answer: (a) 4 : 9
Explanation :
Ratio of volume \( = 8 : 27 \)
\( i.e., \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{8}{27} \)
\( \Rightarrow \frac{r_1^3}{r_2^3} = \frac{8}{27} \)
\( \Rightarrow \frac{r_1}{r_2} = \frac{2}{3} \dots \text{(1)} \)
Ratio of surface area \( = \frac{4\pi r_1^2}{4\pi r_2^2} \)
\( = \left(\frac{r_1}{r_2}\right)^2 \) [From (1)]
\( = \left(\frac{2}{3}\right)^2 = \frac{4}{9} = 4 : 9 \)

Question. A cone of height 24 cm and radius 6 cm at base is made up of clay. A child reshapes it in form of a sphere. The radius of the sphere will be :
(a) 7 cm
(b) 5 cm
(c) 8 cm
(d) 6 cm
Answer: (d) 6 cm
Explanation :
Volume of cone \( = \text{Volume of sphere} \)
\( \Rightarrow \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi R^3 \)
\( \Rightarrow \frac{1}{3} \times \pi \times 6^2 \times 24 = \frac{4}{3}\pi R^3 \)
\( \Rightarrow 6^3 = R^3 \)
\( \Rightarrow R = 6 \text{ cm.} \)

Question. If the surface area of a sphere is 616 cm², then the radius of the sphere is :
(a) 14 cm
(b) 7 cm
(c) 3.5 cm
(d) None of the options
Answer: (b) 7 cm
Explanation :
Surface area of sphere \( = 616 \text{ cm}^2 \)
\( \Rightarrow 4\pi r^2 = 616 \)
\( \Rightarrow 4 \times \frac{22}{7} \times r^2 = 616 \)
\( \Rightarrow r^2 = \frac{616 \times 7}{22 \times 4} \)
\( \Rightarrow r^2 = 49 \)
\( \Rightarrow r = 7 \text{ cm.} \)

Question. A right circular cylinder of radius r cm and height h cm (h > 2r) just encloses a sphere of diameter equal to :
(a) r cm
(b) 2r cm
(c) 3r cm
(d) 4r cm
Answer: (b) 2r cm
Explanation :
Here, maximum diameter of sphere is equal to diameters of cylinder

Question. The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What is the ratio (approximate) of their volumes ?
(a) 1 : 16
(b) 1 : 64
(c) 1 : 4
(d) 1 : 128
Answer: (b) 1 : 64
Explanation :
The diameter of Moon is approximately one-fourth of the diameter of Earth.
Let, Radius of Moon \( = r \),
Then, Radius of Earth \( = 4r \)
Required ratio \( = \frac{\text{Volume of Moon}}{\text{Volume of Earth}} \)
\( = \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi(4r)^3} \)
\( = \frac{r^3}{64r^3} = \frac{1}{64} = 1 : 64 \)

Question. A solid spherical ball of iron of radius 4 cm is melted to form spheres of radius 2 cm each. The number of spheres, so formed is :
(a) 8
(b) 9
(c) 10
(d) 16
Answer: (a) 8
Explanation :
Number of spheres \( = \frac{\text{Volume of a solid sphere with radius 4 cm}}{\text{Volume of a solid sphere with radius 2 cm}} \)
\( = \frac{\frac{4}{3}\pi(4)^3}{\frac{4}{3}\pi(2)^3} = \frac{(2)^6}{(2)^3} = (2)^3 = 8 \)

Question. If the volume of a cube is 729 cm³, what is the length of its diagonal ?
(a) \( 9\sqrt{2} \text{ cm} \)
(b) \( 9\sqrt{3} \text{ cm} \)
(c) 18 cm
(d) \( 18\sqrt{3} \text{ cm} \)
Answer: (b) \( 9\sqrt{3} \text{ cm} \)
Explanation :
Volume of cube \( = (\text{Side})^3 \)
\( 729 = a^3 \Rightarrow a = 9 \text{ cm} \)
Diagonal of cube \( = \text{Side} \times \sqrt{3} \)
\( = 9 \times \sqrt{3} = 9\sqrt{3} \text{ cm} \)

Question. The curved surface area of a right circular cone of radius 14 cm is 440 sq. cm. what is the slant height of the cone ?
(a) 10 cm
(b) 11 cm
(c) 12 cm
(d) 13 cm
Answer: (a) 10 cm
Explanation :
Curved surface area of right circular cone \( = \pi rl \)
\( \Rightarrow 440 = \frac{22}{7} \times 14 \times l \)
\( \Rightarrow l = \frac{440 \times 7}{22 \times 14} = 10 \text{ cm} \)

Question. Two cones have their heights in the ratio of 1 : 3 and radii 3 : 1. The ratio of their volumes is :
(a) 1 : 1
(b) 1 : 3
(c) 3 : 1
(d) 2 : 3
Answer: (c) 3 : 1
Explanation :
Let their heights be x, 3x and their radii be 3y, y.
Then, ratio of volumes \( = \frac{\frac{1}{3}\pi \times (3y)^2 \times x}{\frac{1}{3}\pi \times y^2 \times (3x)} = \frac{9}{3} = 3 : 1 \)

Question. If the volumes of two cones are in the ratio of 1 : 4 and their diameters are in the ratio of 4 : 5, then the ratio of their heights is :
(a) 1 : 5
(b) 5 : 4
(c) 5 : 16
(d) 25 : 64
Answer: (d) 25 : 64
Explanation :
Let the radii of cones be 4x and 5x and their heights be h and H respectively. Then,
\( \Rightarrow \frac{\frac{1}{3}\pi \times (4x)^2 \times h}{\frac{1}{3}\pi \times (5x)^2 \times H} = \frac{1}{4} \)
\( \Rightarrow \frac{h}{H} = \frac{1}{4} \times \frac{25}{16} = \frac{25}{64} \)

Question. A cone of height 7 cm and base radius 3 cm is curved from a rectangular block of wood 10 cm × 5 cm × 2 cm. The percentage of wood wasted is :
(a) 34%
(b) 46%
(c) 54%
(d) 66%
Answer: (a) 34%
Explanation :
Volume of the block \( = (10 \times 5 \times 2) \text{ cm}^3 = 100 \text{ cm}^3 \)
Volume of the cone carved out \( = \left(\frac{1}{3} \times \frac{22}{7} \times 3 \times 3 \times 7\right) \text{ cm}^3 = 66 \text{ cm}^3 \)
Wood wasted \( = (100 - 66) \text{ cm}^3 = 34 \text{ cm}^3 \)
Required % \( = \frac{34}{100} \times 100 = 34\% \)

Question. A cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm. The radius of the cone will be :
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Answer: (d) 8 cm
Explanation :
Let the radius of the cone be r cm.
Volume of cone \( = \text{Volume of cylinder} \)
Then, \( \frac{1}{3}\pi r^2 \times 6 = \pi \times 8 \times 8 \times 2 \)
\( \Rightarrow r^2 = \frac{8 \times 8 \times 2 \times 3}{6} = 64 \)
\( \Rightarrow r = 8 \text{ cm} \)

Question. A cylinder with base radius 7 cm and height 3 cm is melted to form a cone of height 9 cm. Then, radius of the cone is :
(a) 5 cm
(b) \( \frac{7}{2} \text{ cm} \)
(c) 7 cm
(d) 6 cm
Answer: (c) 7 cm
Explanation :
Volume of cylinder \( = \text{Volume of cone} \)
\( \pi r^2 h = \frac{1}{3}\pi R^2 H \)
\( \Rightarrow \pi \times 7 \times 7 \times 3 = \frac{1}{3}\pi \times R^2 \times 9 \)
\( \Rightarrow R^2 = 7^2 \)
\( \Rightarrow R = 7 \text{ cm} \)

Question. The volume of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is : [NCERT Exemplar]
(a) 4 : 3
(b) 9 : 16
(c) 3 : 4
(d) 16 : 9
Answer: (d) 16 : 9
Explanation :
Ratio of volume of two sheres \( = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} \)
\( \Rightarrow \frac{r_1^3}{r_2^3} = \frac{64}{27} \Rightarrow \frac{r_1}{r_2} = \frac{4}{3} \dots \text{(i)} \)
Ratio of surface areas of two sphere \( = \frac{4\pi r_1^2}{4\pi r_2^2} \)
\( = \frac{r_1^2}{r_2^2} = \left(\frac{4}{3}\right)^2 \) [From (i)]
\( = \frac{16}{9} = 16 : 9 \)

Question. A solid sphere of radius 8 cm can be recasted into ........... balls of radius 2 cm.
(a) 40 cm
(b) 64 cm
(c) 12 cm
(d) 10 cm
Answer: (b) 64 cm
Explanation :
Volume of sphere of radius 8 cm \( = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 8^3 \)
Volume of sphere of radius 2 cm \( = \frac{4}{3} \times \pi \times 2^3 = \frac{4}{3}\pi \times 8 \)
No. of balls \( = \frac{\frac{4}{3}\pi \times 8^3}{\frac{4}{3}\pi \times 8} = 8^2 = 64. \)

Question. The total surface area of a hemisphere of radius r is :
(a) πr²
(b) 2 πr²
(c) 3πr²
(d) \( \frac{2}{3} \pi r^2 \)
Answer: (c) 3πr²
Explanation :
Area of hemisphere \( = \text{Area of curved surface} + \text{Area of base} \)
\( = 2\pi r^2 + \pi r^2 = 3\pi r^2 \)

Question. A solid piece of iron of dimensions 49 cm × 33 cm × 24 cm is moulded into a sphere. The radius of the sphere is :
(a) 21 cm
(b) 28 cm
(c) 35 cm
(d) None of the options
Answer: (a) 21 cm
Explanation :
Volume of iron piece \( = (49 \times 33 \times 24) \text{ cm}^3 \)
Let, \( r \) is the radius of the sphere.
Clearly, Volume of sphere \( = \text{Volume of iron piece} \)
\( \frac{4}{3}\pi r^3 = 49 \times 33 \times 24 \)
\( \frac{4}{3} \times \frac{22}{7} \times r^3 = 49 \times 33 \times 24 \)
\( r^3 = \frac{49 \times 33 \times 24 \times 3 \times 7}{4 \times 22} \)
\( r^3 = 49 \times 3 \times 3 \times 3 \times 7 \)
\( \Rightarrow r = 7 \times 3 = 21 \text{ cm} \)

Question. A cylindrical vessel 32 cm high and 18 cm as the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, the radius of its base is :
(a) 12 cm
(b) 24 cm
(c) 36 cm
(d) 48 cm
Answer: (c) 36 cm
Explanation :
Radius of a cylindrical vessel (\( r_1 \)) \( = 18 \text{ cm} \)
and height of a cylindrical vessel (\( h_1 \)) \( = 32 \text{ cm} \)
\ Volume of sand filled in it \( = \pi(r_1)^2 h_1 \)
\( = \pi(18)^2 \times 32 = \pi \times 324 \times 32 \text{ cm}^3 = 10368\pi \text{ cm}^3 \)
Now height of the conical heap (\( h_2 \)) \( = 24 \text{ cm} \)
Let \( r_2 \) be its radius, then \( \frac{1}{3}\pi r_2^2 h_2 = 10368\pi \)
\( \Rightarrow \frac{1}{3} \times r_2^2 \times 24 = 10368 \)
\( \Rightarrow 8\pi r_2^2 = 10368\pi \)
\( r_2^2 = 1296 \Rightarrow r_2 = \sqrt{1296} = 36 \)
Hence, radius of the base of the heap \( = 36 \text{ cm} \)

Question. A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas of the two parts are equal, then the ratio of its radius and the slant height of the conical part is :
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer: (a) 1 : 2
Explanation :
Given, that the radius of the hemisphere and the cone are equal.
Since, the surface areas of the two parts are equal.
\ \( 2\pi r^2 = \pi rl \Rightarrow 2r = l \Rightarrow \frac{r}{l} = \frac{1}{2} \)
So, the ratio 1: 2.

Question. If the surface area of a sphere is \( 616\text{cm}^2 \), then its diameter (in cm) is :
(a) 7
(b) 14
(c) 28
(d) 56
Answer: (b) 14
Explanation :
Surface area of a sphere \( = 616 \text{ cm}^2 \)
\( \Rightarrow 4\pi r^2 = 616 \)
\( \Rightarrow 4 \times \frac{22}{7} \times r^2 = 616 \)
\( \Rightarrow r^2 = 616 \times \frac{7}{22 \times 4} \Rightarrow r = 7 \text{ cm} \)
So, the diameter \( = 2r = 2 \times 7 = 14 \text{ cm.} \)

Question. If each edge of a cube is increased by 50%, the percentage increase in the surface area is :
(a) 50%
(b) 75%
(c) 100%
(d) 125%
Answer: (d) 125%
Explanation :
Let each edge of cube be x.
So, the surface area of the cube \( = 6x^2 \)
Since, the edge of the cube is increased by 50%
\ New edge \( = x + \frac{x}{2} = \frac{3x}{2} \)
So, the new surface area \( = 6\left(\frac{3x}{2}\right)^2 = 6 \times \frac{9x^2}{4} = \frac{27x^2}{2} \)
Increase in the surface area \( = \frac{27x^2}{2} - 6x^2 = \frac{15x^2}{2} \)
Percentage increase \( = \frac{\frac{15x^2}{2} \times 100}{6x^2} = \frac{15}{12} \times 100 = 125\% \)

Question. A right circular cylinder of radius r cm and height h cm (h > 2r) just encloses a sphere of diameter :
(a) r cm
(b) 2r cm
(c) h cm
(d) 2h cm
Answer: (b) 2r cm
Because the sphere encloses the cylinder, therefore, the diameter of sphere is equal to diameter of cylinder which is \( 2r \text{ cm} \).

Question. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is :
(a) 4 cm
(b) 3 cm
(c) 2 cm
(d) 6 cm
Answer: (c) 2 cm
Explanation :
Given, diameter of the cylinder = 2cm
\ Radius = 1 cm
Height of the cylinder = 16 cm
\ Volume of the cylinder = \( \pi r^2 h \)
= \( \pi \times (1)^2 \times 16 \)
= \( 16\pi \text{ cm}^3 \)
Now, let the radius of soild sphere = \( r \text{ cm} \)
Volume of the twelve solid sphere = Volume of cylinder
\( \Rightarrow 12 \times \frac{4}{3} \pi r^3 = 16\pi \)
\( \Rightarrow r^3 = 1 \)
\( \Rightarrow r = 1 \text{ cm} \)
\ Diameter of each sphere, \( d = 2r = 2 \times 1 = 2 \text{ cm} \)
Hence, the required diameter of each sphere is 2 cm.

Question. A cylindrical pencil sharpened at one edge is the combination of :
(a) A cone and a cylinder.
(b) Frustum of a cone and a cylinder.
(c) A hemisphere and a cylinder.
(d) Two cylinders.
Answer: (a) A cone and a cylinder.
Explanation :
Because the shape of sharpened pencil is, a cylinder and a cone.

Question. Volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is :
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) 16 : 9
Answer: (d) 16 : 9
Explanation :
Let the radii of the two spheres are \( r_1 \) and \( r_2 \), respectively.
\ Volume of the sphere of radius, \( r_1 \)
\( V_1 = \frac{4}{3} \pi r_1^3 \dots \text{(i)} \)
and volume of the sphere of radius, \( r_2 \)
\( V_2 = \frac{4}{3} \pi r_2^3 \dots \text{(ii)} \)
Given, ratio of volumes \( V_1 : V_2 = 64 : 27 \)
\( \Rightarrow \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{64}{27} \)
[using Eqs. (i) and (ii)]
\( \Rightarrow r_1^3 : r_2^3 = 64 : 27 \)
\( \Rightarrow r_1 : r_2 = 4 : 3 \dots \text{(iii)} \)
Now, ratio of surface area \( = 4 \pi r_1^2 : 4 \pi r_2^2 \)
\( = r_1^2 : r_2^2 = (4 : 3)^2 \)
[using Eq. (iii)]
Hence, the required ratio of their surface area is 16 : 9.

Question. A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is :
(a) \( 12\pi \text{ cm}^3 \)
(b) \( 15\pi \text{ cm}^3 \)
(c) \( 16\pi \text{ cm}^3 \)
(d) \( 20\pi \text{ cm}^3 \)
Answer: (c) \( 16\pi \text{ cm}^3 \)
Explanation :
A cone is formed be rotating the right angled triangle above the side 3 cm
Height of cone \( (h) = 3 \text{ cm} \)
and radius \( (r) = 4 \text{ cm} \)
\ \( V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times (4)^2 \times 3 \text{ cm}^3 \)
= \( \pi \times 16 = 16\pi \text{ cm}^3 \)

Question. The surface area of a sphere is \( 154 \text{ cm}^2 \). The volume of the sphere is :
(a) \( 179.66 \text{ cm}^3 \)
(b) \( 1500 \text{ cm}^3 \)
(c) \( 1789 \text{ cm}^3 \)
(d) None of the options
Answer: (a) \( 179.66 \text{ cm}^3 \)
Explanation :
Surface area of sphere = \( 154 \text{ cm}^2 \)
\( \Rightarrow 4 \pi r^2 = 154 \)
\( \Rightarrow 4 \times \frac{22}{7} \times r^2 = 154 \)
\( \Rightarrow r^2 = \frac{154 \times 7}{4 \times 22} = \frac{49}{4} \)
\( \Rightarrow r = \frac{7}{2} \text{ cm} \)
Volume of sphere \( = \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 = 179.66 \text{ cm}^3 \)